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Notes of Szemer´edi’s Theorem for

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1 Some notations
In this paper, C will denote various large absolute constants, and c will
denote various small absolute constants.
[a, b] denotes the integers from a to b inclusive. Similarly deﬁne the
[a, b), etc. If E is a set, [E[ will denote the cardinality of E. N will be
0-0
a large integer, and 0 < δ < 1 will be a parameter, and ε = ε(δ) > 0 be
a quantity depending continuously on δ.
Let M = M(N) be a slowly growing function of N such that ε
−C
<
M < N
c
; one could take for instance M = log(N), although one can
get away with a much larger M actually. The exact choice of M is only
important if one is interested in the dependence of N on δ to be chosen
later.
For any function f on Z
N
, deﬁne the Fourier transform
ˆ
f by
ˆ
f :=

m∈Z
N
f(m) exp(2πimn/N),
and deﬁne the Wiener norm |f|
A
∞ by
|f|
A
∞ := |
ˆ
f|
l
∞
(Z
N
)
= sup
n∈Z
N
[
ˆ
f(n)[,
we also deﬁne the modiﬁed A
∞
norm |f|
˙
A
∞
by
|f|
˙
A
∞
:= |
ˆ
f|
l
∞
(Z
N
−{0})
.
We write Y X if Y is not of the form o(X).
0-1
2 The main idea of the proof
Szemer´edi asserts that for any integers k, if N is suﬃciently large de-
pending on δ and k, then every set E ⊂ [0, N] with [E[ δN must
contain a non-trivial arithmetic progression of length k.
We are going to deal with the case k = 4.
The main idea is to show by induction. Let S be the set of all
0 < δ < 1 such that the Szemer´edi theorem hold for k = 4. It is
easy to see that all the δ > 3/4 belong to S, since there is at least one
of a, a + 1, a + 2, a + 3 lies in A, furthermore, we shall prove that if
δ +ε(δ) ∈ S then δ ∈ S.
Assume these, we can give the proof of the theorem.
Proof. Consider the inﬁmum δ
0
of S, now claim δ
0
= 0. If not, δ
0
> 0,
and since δ
0
+ε(δ
0
) ∈ S, we get δ
0
∈ S by induction. By the continuity
of ε, we get
δ +ε(δ) → δ
0
+ε(δ
0
), δ → δ
0
−.
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Therefore there exists δ
1
< δ
0
such that
δ
1
+ε(δ
1
) = δ
0
,
we get δ
1
∈ S by induction, contradict to the inﬁmum of δ
0
.The proof is
done.
Fix δ, let N be a prime suﬃciently large depending on δ and ε, and
let E ⊂ [0, N) be such that [E[ δN, by reﬁning E if necessary we may
assume that
[E[ = δN (1)
we can assume that
[E ∩ P[ (δ +ε)[P[ (2)
for all arithmetic progressions P ⊂ [0, N) of length M by induction.
This is the only place where we shall use the induction hypothesis.
0-3
3 Proof of the induction above
Lemma 3.1. We have
[E ∩ P[ δ[P[ +O(ε[P[) (3)
for all Z
N
arithmetic progression P of length [P[ M
3
Proof. Let r = 0 be the spacing of an arithmetic progression P in Z/NZ.
(i) if 0 < [r[ < N/M. Then we can partition P into the disjoint union of
genuine arithmetic progressions of the same step size r, simply by making
a cut every time the progression wraps around the end of 1, , N.
Except possibly for the ﬁrst and last progression, each of the genuine
progressions has length at least M, and so the density of A in those
genuine progressions is at most δ +ε by (2). Adding up all these density
estimates for those genuine progressions of length at least M, we obtain
the bound
[E ∩ P[ (δ +ε)[P[ + 2M.
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(ii) Assume r is arbitrary. Consider the ﬁrst O(M) multiples of r in Z/NZ.
By the pigeonhole principle, two of them must be < N/M apart, thus
we can ﬁnd a non-zero j = O(M) such that [jr mod N[ < N/M. We
can then partition P into [j[ disjoint progressions, all of length ∼ [P[/[j[,
with step size jr mod N. Applying the previous estimate to all of these
progressions and adding up, we obtain
[E ∩ P[ (δ +ε)[P[ + 2[j[M (δ +ε)[P[ +O(M
2
),
and the claim follows if [P[ is large enough.
We can show that (1) and (3) will imply
|1
E
|
˙
A
∞
= O(εN), (4)
and

k∈Z
N
|1
E∩(E+k)
|
˙
A
∞
2cεN
2
.
For an arbitrary ﬁxed c and N. We can ﬁnd a set B ⊂ Z
N
of size
[B[ cεN such that
[[1
E∩(E+k)
[[
˙
A
∞
cεN.
In fact, let B be the set of all such k that
[[1
E∩(E+k)
[[
˙
A
∞
cεN.
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Hence we have
2cεN
2

[ C
ε
N.
Remark. Balog-Szemer´edi theorem: Let A = ¦a
1
, a
2
, , a
n
¦ be a
ﬁnite set of integers and if ¦(a
i
, a
j
, a
k
, a
l
) ∈ A
4
[a
i
+a
j
= a
k
+a
l
¦ cN
3
,
then there are positive integers c
1
and c
2
and a subset A

j∈(s−P)∩E
exp(2πi(φ
1
(s −j) + (s −j)n(s))/N)[
=O(εN[P[)
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since φ
1
(s−)+(s−)n(s) is a non-constant quadratic function for every
s.
In the case λ = 0 and µ = 0, By a rescaling, we may assume that
the progression P = [0, r). It then suﬃces to show that